You have a flashlight that uses two batteries and you have a package of three new batteries. Suppose that each battery has an independent lifetime (in hours) with exponential(1/50) distribution. Find the distribution of the number of hours T that you can use the
flashlight.
Attempt for solution:
denote by T the number of hours you can use the flash light
let B1, B2 and B3 be the batteries respectively
(B1\B2\B3) ~ exp(1/50)
I have the thought that if we use the first two batteries and the first one runs out first, then we have the second and third one
once the second one runs out we can't use the flash light so the maximum life of the flashlight would be the amount of time the second battery runs
But i don't exactly know how to put this in terms of probability
I would really appreciate if there is someone out there willing to share ideas
Thanks alot!!
Best Answer
So the lifetime of each battery is an exponential distribution $X_{1,2,3}$ with $\lambda=0.02$ when in use and (unlike real batteries) they do not age when not in use.
So put 2 batteries in and the first will fail with $\min\{X_1,X_2\}$ which is exponentially distributed with $\lambda=0.04$.
When the first battery fails, put in battery 3. Because of the memoryless nature of exponential distributions, this is also exponential with $\lambda=0.04$ - but starting at a latter time!
So that gives us the sum of 2 i.i.d. exponential random variables with $\lambda=0.04$. This is an Earlang distribution with $k=2$ and $\lambda=0.04$. So the pdf is:
$$\begin{align}P(T=t) &=\frac{0.04^2te^{-0.04t}}{\Gamma(2)}\\ &=0.016te^{-0.04t}\\ \end{align}$$
See plot.
And the cdf is:
$$\begin{align}P(T\le t) &=\frac{\gamma(2,0.04t)}{(2-1)!}\\ &=\gamma(2,0.04t)\\ \end{align}$$
See plot.